The date on the front of wolframscience.com is true. A
New Kind of Science is at the printer. New sample pages
are available for your perusal, and Stephen picked out a sequence
of new pages that will be shown one per week until the book is in
stores. The pages are not drafts, these are final versions. Page 43
is one of them, showing some of the history of patterns. A 70,000 year
old lattice engraving was discovered too late to be considered
for page 43.

698896 is a palindromic square number with an even
number of digits. The next one is 637832238736. More terms in the series
can be found at the Encyclopedia
of Integer Sequences. A general discussion can be found at worldofnumbers.com.
Toshi "Junk" Kato is intrigued by this sequence, and conjectures
that there are an infinite number of base10 palindromic square numbers
with an even number of digits. He will pay $50 to the first who can
prove him right or wrong. (His money is probably pretty safe.) If
you can make headway on this, let
me know. Vaguely related are the squares
using three digits. In word palindromes, 2002
has just been published, a 2002 word palindromic story.

Toby Gottfried has written a
java applet for my Prime Heart puzzle. I tried solving it myself --
not too hard at all. The New Yorker has written a nice article
about various puzzlers. Stuart Anderson let me know about a very
interesting paper on Squared Squares by Ian
Gambini. It's in french, but has many interesting diagrams:

Juha Hyvönen sent two crossnumber puzzles. In the
first, all entries across and down are square numbers. In the second,
all entries across and down are triangular numbers. Curiously,
both contain all ten digits. Can you find the unique solutions? Send answers.

material added 19 February 2002

When Andrea Gilbert presented her
father's Maze of Life idea, I coined the termed "intelligent cell"
as a modest contribution. Many others have contributed much more. Dr
William Paulsen has been exploring how an intelligent cell can ride
a glider creatively, and has found a sequence (744 steps from Y-start)
that creates a glider gun! It's damn neat, watching an intelligent cell
manipulate it's universe. (To watch it, undo all, then click
through redo.) Andrea has also added new Plank maze puzzles at her
site (clickmazes.com).

Serhiy Grabarchuk of Puzzles.com
recently answered a question about folding puzzles. He was so
comprehensive, I felt obligated to quote his paperfolding
list.

There is a group of people competing for who can
compute a million digits of Pi the fastest on an average PC. Stu's Pi Page
lists the various links and records. His program can produce a meg
of Pi in less than 9 seconds.

I've updated quite a few previous puzzles to link to a
solvers list and solutions.

Erich Friedman: Here are a set of fifth powers with an
odd property: A=27^5=14348907, B=39^5=90224199,
C=1239^5=2919823044470199. C is composed of the same digits as A and
B together. What is the smallest set of fourth powers with this
property?

material added 5 February
2002

Siam News has
published a $100,
hundred digit contest by Nick Trefethen of Oxford University. There
are lots of interesting problems here. If you think you can solve one
of these, and would like to be a part of a Mathpuzzle team, let me know. I'll hook you
up with a small group.

The
Zen of Magic Squares, Circles, and Stars by Clifford Pickover. During
the 2000 National Puzzler's
League convention, octogenarian member Twisto shared with us
some of the magnificent constructions of The Great Fubine, a man who
turned to magic squares after losing all his money in the stock market
crash of 1929. Twisto told me about how Fubine was mostly ignored by
mathematicians of the time -- but he thought the material was
deserving of attention. I mentioned this to Cliff. He followed up
big time: first by visiting Twisto at his home, and then by
publishing it. Cliff uncovered all sorts of things while
researching this book. Good stuff -- if you like magic squares, this
is a book you need. And, if you don't ... well, there's lots of
magic squares in here. See if you can find it at your bookstore,
at the very least. Another recent book on the same topic is Magic
Square Lexicon: Illustrated, by Harvey
D. Heinz and John
R. Hendricks.

Martin Gardner frequently debunks myths and explains scams. One that
has annoyed me lately is known as The Spanish Prisoner scam. Fabulous
wealth is promised, for nothing. I've gotten three different letters in the
past two weeks alone. It turns out that these are far nastier than
I first believed. After all, I thought it was just a scam. The United State
Secret Service explains the full horror. (I have forwarded all of
these letters to the Secret Service, for their perusal.) Hoaxbusters is another good
page explaining email hoaxes and scams, so is Snopes.

Dave Tuller sent me this: Given an arbitrary polynomial with
rational coefficients f(x), show that you can always find another
polynomial with rational coefficients g(x) so that f(x)g(x) is a
linear combination of prime powers of x. (a x^2 + b x^3 + c x^5 +
d x^7 + ..., with rational {a,b,c,d,...}, and all powers of x
prime) For example, 1+x works since x^2(1+x)=x^2+x^3. Send Answer.

Erich Friedman's 1-24 square challenge ... which I offered $10 for
... has been solved by Erich Friedman. I'll write up solvers and more
things on Friday.

material added 28 January
2002

This week's entry requires a bit of work for some, but
the results are well worth it. Many of the best mathematical files
are in a ps.gz format. This first paragraph is aimed at Windows
users that don't already have access to files of this type. First,
you'll need to get AFPL Ghostscript 7.03 (Instructions,Download).
Once you've installed that, you can get Ghostview (Instructions,Download).
If you're on a Linux or Unix system, you likely have both already.

Now that you're equipped to look at the files, Don
Knuth has released the first
part of the long awaited Book 4. It is two parts: Generating
all n-tuples and Generating
all permutations. These are fantastic articles, well worth
perusing. One piece in these I'd like to fix involves the Chinese
Rings puzzle. Somewhere(1, 2,3, 4), I read
that Cardano
had described these in one of his books. UIUC just happened to have an
original copy of that particular 15th century book, and the chinese
rings weren't in it. I'd like to find an original source, for
Knuth's book. Can anyone help? This would imply an early
understanding of an important math concept, by means of a puzzle.

Helmut Postl did an exploration of the Tetrods, and
found that three identical shapes like those below could be made.
The holes are in the same place, too -- where are they? He found a mirror symmetric solution for the
Iamodds. Working on my 1.9 * fourthroot(9.1) = 3.30000007 problem,
he noticed: 19 * fourthroot(9.1) :: 33, 19^44 * 9.1 :: 33^4, 19^4 *
91 :: 33^4 * 10, 11859211 :: 11859210.

Shyam
Sunder Gupta has a great new website on number recreations. An
example from his site: find three consecutive abundant numbers.

Erich Friedman: It is well known
that 1^2 + 2^2 + ... + 24^2 = 70^2. Can we tile a 140x140 square with 4
copies of each square of size 1 through 24? I'll offer $10 for an
answer to this.

Nick Baxter has started a 1000 Play
Thinks Errata page. On item 150, many trees can all be touching each
other, and thus be equidistant.

Is there always a prime between two square numbers?
This is a famous unsolved math problem, so I took a quick look at
it. In Mathematica, PrimePi[x] gives the number of primes
less than or equal to x. The part in blue thus counts the number of
primes between n^2 and (n+1)^2. The part in green is a lower bound
I concocted. Up to n=100000, at least, it works pretty well.

material added 14 January
2002

M Oskar van Deventer has made a
made a wonderful challenge he calls the Harbour
puzzle. (I can't seem to get the last few blocks into the ship.)

A lot of people have recently
recommended 1000
Play Thinks by Ivan Moscovich to me: Nick Baxter, Will Shortz, and Ken Duisenberg.
I also liked it. I spotted one error: In Playthink 150, Ivan shows how 3
trees can be equidistant (in a triangle) and asks if that is the highest
number of equidistant trees. His answer is wrong. I would recommend
looking in a bookstore for it, first. I found many copies in every
bookstore I tried, and this is a huge book. So, save on your
shipping costs. Any corrections you find should be emailed to Nick
Baxter at nick@baxterweb.com.

Another marvelous book is Puzzlers'
Tribute, a hardbound synopsis of Gathering
for Gardner 4. Almost every famous recreational mathematician I
know has a short, interesting paper inside it. An online version of
the previous book is available for free at the site. I can hardly
wait to see the next book in this series. Sadly, this book doesn't
seem to have caught the attention of bookstore chains, and it
really deserves it.

At some point during the week, I wondered if a grid
could be divided into pairs of points where every distance was
distinct. It turns out that this is possible on a 10x10 grid. This
makes a marvelous little solitaire exercise. I found the answer
below by hand. The two 5's
are separated by the distance Sqrt[5], the 13's
by Sqrt[13], and so on. Every possible distance in a 10x10 grid is
represented.

For the puzzle of the week I'll
defer to the NPR puzzle of the
week. Quoting Will Shortz: "This is a timely challenge from
listener Ed Pegg, Jr. of Champaign, Illinois. He points out that 2002
is an unusual number because it's a palindromic number (it reads
forward and backward the same). Now double that: 20022002. The
challenge is to find three different palindromic numbers that can
be multiplied together to get the palindromic number 20,022,002.
Each of these numbers must have at least 2 digits. What numbers are
they?" You can enter the challenge at the National Public
Radio site.

I discovered the above while concocting a different
problem. I've seen two palindromic years so far: 1991 and 2002. If you
multiply those together, you get 3985982, which isn't palindromic.
Few consecutive palindromic number sets have a palindromic product.
101 × 111 × 121 = 1356531 is one example. Find the next
triple of palindromic numbers with a palindromic product. Answer.

The above plot shows Power[Apply[Times,
ContinuedFraction[k,n]],1/n], for various k. I picked p, 2p, g, Log[2], 2^(1/3), 3^(1/3). As n increases, the
geometric mean of all of these numbers converge to the Khinchin
constant (blue line). That's pretty neat.

As something of an ultimate "oops", Dr. Robert Smith
looked at the well-accepted "Earth eventually gets engulfed by expanding
Sun" argument, and noticed that no-one had bothered to account for Sun
mass being converted to energy. It turns out that the Earth will not
suffer a firey doom within a red giant inferno! Huzzah! As the sun loses
mass, the Earth will spiral out to a orbit beyond the reach of the
eventual red giant, 7 billions years from now. Here
is the report. Just goes to show that it never hurts to check a
calculation. I might be celebrating too soon, though -- our galaxy will
collide with another in about 3 billion years.

Scott Kim did a lovely article on Knot Puzzles in the
latest Discover magazine. He mentions the Picture Hanging
problem.

Xort[A_List] :=
Fold[Join[Reverse[Take[##]],Drop[##]]&, A,
Flatten[Position[BitXor@@@Partition[Append[A,1],2,1],1]]] is an
interesting method for sorting pancakes of two different sizes. In the
picture, a green line indicates same-size/no-flop, and a red line
indicates different/flop. A computer uses Xor to tell if two bits
or the same. This can be called the Pancake Sort, or perhaps just
Xort.

Pancake Xorting

In other words, a spatula goes down a stack of
pancakes, and flops the pancakes above it every time the sizes
differ. When the spatula reaches the last pancake, the stack is
flopped so that larger pancakes are on the bottom. This is a
reversible sorting technique. By saving the BitXor list, the
original state can be reconstructed. Larger integers can be sorted
by considering each bit in a series. Can the BitXor lists be combined in
any interesting way to sort Pancake stacks in an optimized way? Send answer. Many people responded to Eric's pancake problem from
last week. Turns out this was at Sloane's
Integer Sequences, but there was a typo in the sequence there.
jkmclean pointed out the interesting fact that the sequence could
increase by no more than 2.

Peter Esser found a lovely solution for a particular
type of polyform:

Eric Iverson: 2002 has the property that the product of
the non-zero digits equals their sum. What was the last year in which
this happened, and the next?

Erich
Friedman: I couldn't let the year end without making a new years
puzzle. Start with the annulus of sqaures shown below. Place 4
horizontal dominoes (each covering two adjacent squares) so that the
remaining squares can be tiled with dominoes in exactly 2002 ways.
I think there's only one solution (up to symmetry). Have a happy
new year.

Several people sent answers to Theo Gray's Spieker Center Table challenge. Only Dick
Saunders Jr. and I solved Serhiy Grabarchuk's Constellation puzzle
(he runs Puzzles.com). It's
very nice, so I'm repeating it, hoping more people will try to solve
it. Answers.

material added 27 December
2001

Eric Weisstein and I chatted a bit about the Pancake
sorting problem. If you have a spatula, and 7 pancakes of size 1-7
on a plate, what is the minimum number of flops you'll need to get the
pancakes sorted from smallest to largest? Eric took a brute force
look at the problem. 2 pancakes - 1 flop. 3 pancakes - 3 flops {1,3,2}.
4 pancakes - 4 flops {2, 4, 1, 3}. 5 pancakes - 5 flops {1, 3, 2, 5,
4}. 6 pancakes --- 7 flops. There are only two "worst case" stacks:
{4, 6, 2, 5, 1, 3} and {5, 3, 6, 1, 4, 2}. Below is the method for
sorting out the first stack. Can you you sort the second stack with
7 flops? Answer. The worst case for 7
pancakes is unknown, so far as I know. Can someone analyze this?
The sequence so far is 1 3 4 5 7, if you can extend this, let me know.

Eric's worst 6 pancake stack needs 7 spatula flops to sort.

I've recently been studying the possibilities of Domino
Poker. No real sane reason - it just struck me as something that needed
to be investigated, and it hasn't left me alone since. I love the
feel of a domino set. I'd absolutely love to have a mini double-9
set. Suppose I draw 5 dominoes at random. I might get something
like , or 6-6, 6-2, 4-4,
5-1, 4-5. Looking at the distribution of numbers, I've gotten 2
singles, a double, and 2 triples. Or {1~1~2~3~3}. There are 28!/23!/5!
= 98280 possible hands with the 28 double-6 dominoes. Here is an
analysis of all possible distributions: 105={2~4~4}, 105={3~3~4},
105={1~1~1~1~6}, 420={1~1~3~5}, 420={1~1~4~4}, 420={1~2~2~5},
420={1~1~1~1~1~1~4}, 462={1~1~1~1~1~5}, 798={2~2~2~2~2}, 840={1~3~3~3},
840={2~2~2~4}, 1680={2~2~3~3}, 1680={1~1~1~2~5}, 3360={1~2~3~4},
4200={1~1~1~3~4}, 4305={1~1~1~1~3~3}, 4620={1~1~1~1~1~2~3},
5250={1~1~1~1~2~2~2}, 6090={1~1~1~1~2~4}, 7980={1~1~2~2~2~2},
8190={1~1~2~2~4}, 10920={1~2~2~2~3}, 11970={1~1~2~3~3},
23100={1~1~1~2~2~3}. The rarest hands have only 3 different numbers or
6 of a kind. My random hand was the second most common. There are
many other hands to consider. For example, there are straights and
other things. is a
closed chain where all the connecting sums have 8 pips. Many different
letters can be formed, like A or O.

Theo Gray, a Mathematica front-end developer,
recently built a table. And not just any table -- it's a full-blown
3-4-5 Spieker Circle triangle table, with inlays. He kept a photo diary of his effort.
Eric Weisstein has added this to his Spieker
Circle description. Eric and I both helped a tiny bit, mainly
by pointing to Clark Kimberling's excellent Encyclopedia
of Triangle Centers, and by perusing CK's Triangle
Book (one of my favorites). Theo came up with the built in puzzle,
which is my puzzle of the week. ABC is a 3-4-5 triangle, with
midpoints DEF. Prove that the incenter of triangle ABC is on
the incircle of DEF, as shown below. Answer.

I absolutely adored issue 56 of Cubism For Fun. Every page of it.
In particular, I enjoyed a write-up about Alex Randolf's "Tetrods".
He took the 5 tetrominoes, and drilled a single hole in each in all
possible ways to make 12 pieces. Then he tossed in a 1x1 square with
a hole. The total area of these pieces is 49. It is quite easy to
make a 7x7 square. Trickier is to make a 7x7 square with 4-fold
symmetry. If the 1x1 piece is left out, can 3 identical shapes be
made? 4? Here's the same sort of idea with Iamonds ... I suppose
I'll call these Iamodds.

Puzzle of the Week: Small decimals have just one
digit on each side of the dot. 3.3 is a small decimal. There is one
small decimal with a curious property. If you multiply its reverse
and its fourth root, you get a number remarkably close to 3.3
. So, for example, if I start with 1.3, hit the square root button
twice, and multiply by 3.1, I get 3.31. That's kind of close, but
another small decimal is remarkably closer. What is it? Many people
sent me the Answer. Eric Weisstein
has a page devoted to Almost
Integers.

The big new Mersenne prime from 5 December, and Richard
Schroeppel's posting of DN Lehmer's classic factoring story, "Hunting Big Game
in the Theory of Numbers" inspired me to examine primes. I
wondered what the lowest prime factor of (10^(10^7))! + 1 might be. Such
a prime would have at least ten million digits. Mathematica
beckoned to me to do a little study of the lowest prime factors of
various factorials +1, so I took a look via Table[{n,
FactorInteger[(n-1)! + 1, FactorComplete -> False][[1, 1]]}, {n,
1, 100}]. I did a bit of clean up on the data by entering simpler
forms for factorial primes like 77!+1, and by substituting xx for
numbers Mathematica couldn't factor quickly. Record
factorial primes discussed here.

In the data below, note how the smallest factor and n
are the same whenever n is prime. This is known as Wilson's
Theorem. Perusing the data, I noticed a different sequence:
{4,7}, {6,11}, {10,19}, {22,43}, {24,47}, and so on. Wilson basically
asked about p|((p-1)!+1), where "|" indicates
divisibility. Does p divide into (p-1)!+1, in other
words. (2|6 is true, since 6/2 is an integer.)

Here is a plot of the gaps between the numbers on the
first list. It turns out this is an aspect of Wilson's
Corollary. The original Wilson Theorem is that p|((p-1)!+1)
for any prime number p. A lesser known result is that p|((p-2)!-1)
for any prime number p. What I rediscovered is that for any
prime of the form 4k-1 divides ((2k-1)!)^2 -1. This
implies (4k-1)|((2k-1)!+1) or (4k-1)|((2k-1)!-1),
if (4k-1) is prime. Which one? I don't know. I'll pay $5
to the first person that can send me a formula for it. I made a plot
of the progressive +1 and -1 terms for the 4k-1 primes and it's
pretty chaotic. I also noticed that for primes of the form 4k+1,
(4k+1)|(((2k-1)!)^2 + 4).

I thought the above might lead to a good puzzle, but I
managed to solve it. I learned something in the process.

Lloyd King has started a new puzzle site
with some nice ideas. Andrea Gilbert
has started analyzing Denki Block
puzzles. These are kind of like sliding magnet puzzles. Andrea has
found a very difficult puzzle in a 4x4 grid.

Zome Systems
has updated their site. The Big Bag of Parts is a great deal.

In response to my message from 5 December, Ron Zeno
took a look at sorting 10 elements, and found an improvement over
the best known network (Waksman, 1969), which had length 29, delay
9. Zeno found a lot of length 29, delay 8 networks. Ron Zeno:
"Waksman did the real work, I just analyzed the solution. There are
literally over a million equivalent networks: 1. Any set of 5
disjoint pairs works for the first 5 comparators (945 possibilities),
2. Any 8 of the 10 elements can be used for the next 4 comparators (45
possibilities, and there are 3 or so arrangements of the 8 chosen that
will work), 3. After that, there are at least 8 variations of the
remaining network = 945*45*3*8 = 1,020,600. Lots of solutions in a
vast solution space. (about 9,495^8 different networks of delay 8.
That makes the chance of randomly finding one is about 1 in 6.6 *
10^26)." Here is Zeno's list of pairs: {{1, 2}, {3,
4}, {5, 6}, {7, 8}, {9, 10}, {1, 3}, {2, 4}, {5, 7}, {6, 8}, {1, 9},
{2, 7}, {8, 10}, {1, 5}, {3, 8}, {4, 10}, {6, 9}, {2, 6}, {3, 5}, {4,
9}, {7, 8}, {2, 3}, {4, 7}, {5, 6}, {8, 9}, {3, 5}, {4, 6}, {7, 8},
{4, 5}, {6, 7}}. Ron noticed that another 8-delay network
can be gotten just by starting with Waksman's original sorter and
shifting the last operation back two positions.Ron- "It's too
trivial a change to not have been found by Waksman." Ron also
pointed me to a published paper for 9 elements, I. Parberry, "A
Computer Assisted Optimal Depth Lower Bound for Nine-Input Sorting
Networks", Mathematical Systems Theory, Vol. 24, pp. 101-116,
1991. A second paper by Parberry is Parberry: "On
the Computational Complexity of Optimal Sorting Network Verification".
Proceedings of The Conference on Parallel Architectures and Languages
Europe, Springer-Verlag Lecture Notes in Computer Science, Vol. 506, pp.
252-269, June 1991.

Waksman's 9-delay sorting network

Ron Zeno's 8-delay sorting network

Tom Thomas, Roosevelt University: "Bob Kraus suggested
that I contact you about a search that I am doing with a high school
math teacher. We are looking for logic problems that come from different
cultures, e.g., from Asia, Africa that have been translated into
English. We are trying to emphasize the international interest in
logic and felt that providing high school students with a sampler
of logic puzzles that come from different parts of the world would
help to make this point. Do you know of any resource that we can
consult that would include such information?" If can help here,
please send an email to Tom
Thomas.

Don Woods: "I write with some embarrassment this time.
I noticed that you had posted some answers to my 20 Questions puzzle,
and when I looked at them they didn't match the answer I had in mind,
yet (unlike some I'd seen so far) they did appear to be correct.
After some effort I found the bug in the program I'd written that
"proved" the uniqueness of my intended solution. Oops! Sheer bad
luck that the bug happened to be such that the one solution it did
report was the one I expected, instead of some other. Sigh.With
some effort, I've managed to swap a few questions around and tweak
a couple answers here and there, and believe I have "uncooked" the
puzzle. I enclose it here." If you
can solve 20 Q, Mark 2, write me.

I just so happen to be playing Don's classic game
Adventure on my Visor Edge (which just had a $50 drop in price). I've
been programming in Lua and C on the thing. I'll try to have a
rundown on neat Palm OS software in my next update.